CorrectionKey=NL-A;CA-A CorrectionKey=NL-C;CA-C 3 . 3 DO

© H

oug

hton

Mif

flin

Har

cour

t Pub

lishi

ng

Com

pan

y

Name Class Date

Resource Locker

Explore Investigating Real Solutions of Quadratic Equations

A Complete the table.

a x 2 + bx + c = 0 a x 2 + bx = -c f (x) = a x 2 + bx g (x) = -c

2 x 2 + 4x + 1 = 0

2 x 2 + 4x + 2 = 0

2 x 2 + 4x + 3 = 0

B The graph of ƒ (x) = 2 x 2 + 4x is shown. Graph each g (x) . Complete the table.

Equation Number of Real Solutions

2 x 2 + 4x + 1 = 0

2 x 2 + 4x + 2 = 0

2 x 2 + 4x + 3 = 0

C Repeat Steps A and B when ƒ (x) = -2 x 2 + 4x.

a x 2 + bx + c = 0 a x 2 + bx = -c f (x) = a x 2 + bx g (x) = -c

-2 x 2 + 4x - 1 = 0

-2 x 2 + 4x - 2 = 0

-2 x 2 + 4x - 3 = 0

Equation Number of Real Solutions

-2 x 2 + 4x - 1 = 0

-2 x 2 + 4x - 2 = 0

-2 x 2 + 4x - 3 = 0

0

y

-2-4 2

2

4

x

0

y

-2-4

2

-24

x

2 x 2 + 4x = -1

2 x 2 + 4x = -2

2 x 2 + 4x = -3

f (x) = 2 x 2 + 4x g (x) = -1

g (x) = -2

g (x) = -3

f (x) = 2 x 2 + 4x

f (x) = 2 x 2 + 4x

2

1

0

-2 x 2 + 4x = 1

-2 x 2 + 4x = 2

-2 x 2 + 4x = 3

f (x) = -2 x 2 + 4x g (x) = 1

g (x) = 2

g (x) = 3

f (x) = -2 x 2 + 4x

f (x) = -2 x 2 + 4x

2

1

0

Module 3 139 Lesson 3

3.3 Finding Complex Solutions of Quadratic Equations

Essential Question: How can you find the complex solutions of any quadratic equation?

DO NOT EDIT--Changes must be made through “File info” CorrectionKey=NL-A;CA-A

A2_MNLESE385894_U2M03L3.indd 139 19/03/14 12:00 PM

Common Core Math StandardsThe student is expected to:

N-CN.C.7

Solve quadratic equations with real coefficients that have complex solutions. Also N-CN.C.2, A-REI.B.4b

Mathematical Practices

MP.2 Reasoning

Language ObjectiveWork with a partner or small group to determine whether solutions to quadratic equations are real or not real and justify reasoning.

COMMONCORE

COMMONCORE

HARDCOVER PAGES 99108

Turn to these pages to find this lesson in the hardcover student edition.

Finding Complex Solutions of Quadratic Equations

ENGAGE Essential Question: How can you find the complex solutions of any quadratic equation?Possible answer: You can factor, if possible, to find real solutions; approximate from a graph; find a square root (which may be part of completing the square); complete the square; or apply the quadratic formula. For the general equation a x 2 + bx + c = 0, you must either complete the square or use the quadratic formula to find the complex solutions of the equation.

PREVIEW: LESSON PERFORMANCE TASKView the Engage section online. Discuss the photo and how to solve a quadratic equation to determine how high a baseball will go after it is hit. Then preview the Lesson Performance Task.

139

HARDCOVER

Turn to these pages to find this lesson in the hardcover student edition.

© H

ough

ton

Mif

flin

Har

cour

t Pub

lishi

ng C

omp

any

Name

Class Date

Resource

Locker

Explore Investigating Real Solutions of Quadratic Equations

Complete the table.

a x 2 + bx + c = 0a x 2 + bx = -c

f (x) = a x 2 + bx g (x) = -c

2 x 2 + 4x + 1 = 0

2 x 2 + 4x + 2 = 0

2 x 2 + 4x + 3 = 0

The graph of ƒ (x) = 2 x 2 + 4x is shown. Graph each g (x) . Complete the table.

EquationNumber of Real

Solutions

2 x 2 + 4x + 1 = 0

2 x 2 + 4x + 2 = 0

2 x 2 + 4x + 3 = 0

Repeat Steps A and B when ƒ (x) = -2 x 2 + 4x.

a x 2 + bx + c = 0a x 2 + bx = -c

f (x) = a x 2 + bx g (x) = -c

-2 x 2 + 4x - 1 = 0

-2 x 2 + 4x - 2 = 0

-2 x 2 + 4x - 3 = 0

EquationNumber of Real

Solutions

-2 x 2 + 4x - 1 = 0

-2 x 2 + 4x - 2 = 0

-2 x 2 + 4x - 3 = 0

N-CN.C.7 Solve quadratic equations with real coefficients that have complex solutions.

Also N-CN.C.2, A-REI.B.4bCOMMONCORE

0

y

-2-42

2

4

x

0

y

-2-4

2

-2

4

x

2 x 2 + 4x = -1

2 x 2 + 4x = -2

2 x 2 + 4x = -3

f (x) = 2 x 2 + 4x g (x) = -1

g (x) = -2

g (x) = -3f (x) = 2 x 2 + 4x

f (x) = 2 x 2 + 4x

2

1

0

-2 x 2 + 4x = 1

-2 x 2 + 4x = 2

-2 x 2 + 4x = 3

f (x) = -2 x 2 + 4x g (x) = 1

g (x) = 2

g (x) = 3f (x) = -2 x 2 + 4x

f (x) = -2 x 2 + 4x

2

1

0

Module 3

139

Lesson 3

3 . 3 Finding Complex Solutions

of Quadratic Equations

Essential Question: How can you find the complex solutions of any quadratic equation?

DO NOT EDIT--Changes must be made through “File info”

CorrectionKey=NL-A;CA-A

A2_MNLESE385894_U2M03L3.indd 139

19/03/14 12:05 PM

139 Lesson 3 . 3

L E S S O N 3 . 3

DO NOT EDIT--Changes must be made through “File info”CorrectionKey=NL-C;CA-C

© H

oug

hton Mifflin H

arcourt Publishin

g Com

pany

Reflect

1. Look back at Steps A and B. Notice that the minimum value of f(x) in Steps A and B is -2. Complete the table by identifying how many real solutions the equation ƒ (x) = g (x) has for the given values of g(x).

2. Look back at Step C. Notice that the maximum value of ƒ (x) in Step C is 2. Complete the table by identifying how many real solutions the equation ƒ (x) = g (x) has for the given values of g (x) .

3. You can generalize Reflect 1: For ƒ (x) = a x 2 + bx where a > 0, ƒ (x) = g (x) where g (x) = -c has real solutions when g (x) is greater than or equal to the minimum value of ƒ (x) . The minimum value of ƒ (x) is

ƒ (- b _ 2a ) = a (- b _ 2a ) 2 + b (- b _ 2a ) = a ( b 2 _ 4 a 2

) - b 2 _ 2a = b 2 _ 4a - b 2 _ 2a = b 2 _ 4a - 2 b 2 _ 4a = - b 2 _ 4a .

So, ƒ (x) = g (x) has real solutions when g (x) ≥ - b 2 _ 4a .

Substitute -c for g (x) . -c ≥ - b 2 _ 4a

Add b 2 __ 4a to both sides. b 2 _ 4a - c ≥ 0

Multiply both sides by 4a, which is positive. b 2 - 4ac ≥ 0

In other words, the equation a x 2 + bx + c = 0 where a > 0 has real solutions when b 2 - 4ac ≥ 0.

Generalize the results of Reflect 2 in a similar way. What do you notice?

Value of g (x) Number of Real Solutions

off (x) = g (x)

g (x) = -2

g (x) > -2

g (x) < -2

Value of g (x) Number of Real Solutions

off (x) = g (x)

g (x) = 2

g (x) > 2

g (x) < 2

1

1

2

0

0

2

For f (x) = a x 2 + bx where a < 0, f (x) = g (x) where g (x) = -c has real solutions when g (x) is

less than or equal to the maximum value of f (x) . The maximum value of f (x) is f (- b ___ 2a

) = - b 2 ___ 4a

.

So, f (x) = g (x) has real solutions when g (x) ≤ - b 2 ___ 4a

.

Substitute -c for g (x) . -c ≤ - b 2 __ 4a

Add b 2 ___ 4a to both sides. b 2 __ 4a - c ≤ 0

Multiply both sides by 4a, which is negative. b 2 - 4ac ≥ 0

Whether a > 0 or a < 0, b 2 - 4ac ≥ 0 tells when a x 2 + bx + c = 0 has real solutions.


DO NOT EDIT--Changes must be made through “File info” CorrectionKey=NL-B;CA-B


A2_MNLESE385894_U2M03L3 140 5/22/14 10:47 AM

EXPLORE Investigating Real Solutions of Quadratic Equations

INTEGRATE TECHNOLOGYStudents can use a graphing calculator to graph f (x) and each function g (x) to verify the number of real solutions to each equation.

QUESTIONING STRATEGIESIf an equation is written in vertex form, what information can you use to find out if it has

real solutions? The sign of a determines the direction of the opening and the maximum or minimum value tells you whether there are real solutions.

How do you determine where the graph of a quadratic function crosses the x-axis? You

can find the x-intercepts of the graph of a quadratic function in standard form by factoring the function to get its intercept form. If the function is not factorable, the x-intercepts can be found by using the quadratic formula to find the zeros of the function.

PROFESSIONAL DEVELOPMENT

Math BackgroundIn Algebra 1, students used the quadratic formula to find real solutions to a quadratic equation. Students now revisit the formula to extend its use to complex solutions.

The sign of the expression b 2 - 4ac determines whether the quadratic equation has two real solutions, one real solution, or two nonreal solutions. For cubic equations of the form a x 3 + b x 2 + cx + d = 0, the sign of the discriminant b 2 c 2 - 4a c 3 - 4 b 3 d - 27 a 2 d 2 determines whether the equation has three real solutions, two real solutions, or one real solution.

Finding Complex Solutions of Quadratic Equations 140


© H

oug

hton

Mif

flin

Har

cour

t Pub

lishi

ng

Com

pan

y

Explain 1 Finding Complex Solutions by Completing the Square

Recall that completing the square for the expression x 2 + bx requires adding ( b __ 2 ) 2 to it, resulting in the perfect square

trinomial x 2 + bx + ( b __ 2 ) 2 , which you can factor as (x + b __ 2 ) 2 . Don’t forget that when x 2 + bx appears on one side of an

equation, adding ( b __ 2 ) 2 to it requires adding ( b __ 2 ) 2 to the other side as well.

Example 1 Solve the equation by completing the square. State whether the solutions are real or non-real.

3 x 2 + 9x - 6 = 0

1. Write the equation in the form x 2 + bx = c.

3 x 2 + 9x - 6 = 0

3 x 2 + 9x = 6

x 2 + 3x = 2

2. Identify b and ( b __ 2 ) 2 . b = 3

( b _ 2 ) 2 = ( 3 _ 2 ) 2 = 9 _ 4

3. Add ( b __ 2 ) 2 to both sides of the equation.

x 2 + 3x + 9 _ 4 = 2 + 9 _ 4

4. Solve for x.

(x + 3 _ 2 ) 2 = 2 + 9 _ 4

(x + 3 _ 2 ) 2 = 17 _ 4

x + 3 _ 2 = ± ―― 17 _ 4

x + 3 _ 2 = ± √_ 17 _ 2

x = - 3 _ 2 ± √_ 17 _ 2

x = -3 ± √_ 17 _ 2

There are two real solutions: -3 + √_ 17 _ 2

and -3 - √_ 17 _ 2 .

x 2 - 2x + 7 = 0

1. Write the equation in the form x 2 + bx = c.

2. Identify b and ( b __ 2 ) 2 . b =

( b _ 2 ) 2 = ( _ 2 ) 2 =

3. Add ( b __ 2 ) 2 to both sides.

x 2 - 2x + = -7 +

4. Solve for x.

x 2 + 2x = -7 +

(x - ) 2 =

x - = ±

――

x = 1 ±

――

There are two real/non-real solutions:

and .

x 2 - 2x = -7

-2

-21

11

11

-6

-6

-6

1

1

1 + i √ ― 6

1 - i √ ― 6


DO NOT EDIT--Changes must be made through “File info”CorrectionKey=NL-A;CA-A

A2_MNLESE385894_U2M03L3.indd 141 19/03/14 12:00 PM

COLLABORATIVE LEARNING

Peer-to-Peer ActivityHave students work in pairs. Provide each pair with several quadratic equations written in various forms. Have one student verbally instruct the partner in how to find the nonreal solutions to the equation. Then have partners switch roles, repeating the activity for a different quadratic equation. Have students discuss how their steps for solving the equation were similar or different.

EXPLAIN 1 Finding Complex Solutions by Completing the Square

QUESTIONING STRATEGIESHow do you convert quadratic functions to vertex form? Explain. You can convert

quadratic functions from standard form to vertex form f (x) = a (x - h) 2 + k by completing the square on a x 2 + bx. You have to add and subtract the same constant to keep the function value the same.

INTEGRATE MATHEMATICAL PRACTICESFocus on TechnologyMP.5 Discuss with students how to use the graphing calculator to find a maximum or minimum value of a quadratic function. Students can solve problems algebraically and then use their graphing calculators to check their solutions.

141 Lesson 3 . 3


© H

oug

hton Mifflin H

arcourt Publishin

g Com

pany • Im

age C

redits: ©

Fred Fokkelm

an/Shuttersto

ck

Reflect

4. How many complex solutions do the equations in Parts A and B have? Explain.

Your Turn

Solve the equation by completing the square. State whether the solutions are real or non-real.

5. x 2 + 8x + 17 = 0 6. x 2 + 10x - 7 = 0

Explain 2 Identifying Whether Solutions Are Real or Non-realBy completing the square for the general quadratic equation a x 2 + bx + c = 0, you can obtain the quadratic

formula, x = -b ± √__

b 2 - 4ac ___________ 2a , which gives the solutions of the general quadratic equation. In the quadratic formula, the expression under the radical sign, b 2 - 4ac, is called the discriminant, and its value determines whether the solutions of the quadratic equation are real or non-real.

Value of Discriminant Number and Type of Solutions

b 2 - 4ac > 0 Two real solutions

b 2 - 4ac = 0 One real solution

b 2 - 4ac < 0 Two non-real solutions

Example 2 Answer the question by writing an equation and determining whether the solutions of the equation are real or non-real.

A ball is thrown in the air with an initial vertical velocity of 14 m/s from an initial height of 2 m. The ball’s height h (in meters) at time t (in seconds) can be modeled by the quadratic function h (t) = -4.9 t 2 + 14t + 2. Does the ball reach a height of 12 m?

Set h (t) equal to 12. -4.9 t 2 + 14t + 2 = 12

Subtract 12 from both sides. -4.9 t 2 + 14t + 10 = 0

Find the value of the discriminant. 14 2 - 4 (-4.9) (-10) = 196 - 196 = 0

Because the discriminant is zero, the equation has one real solution, so the ball does reach a height of 12 m.

Each equation has two complex solutions, because the set of complex numbers includes

all real numbers as well as all non-real numbers.

x 2 + 8x = - 17 x 2 + 8x +16 = - 17 + 16 (x + 4) 2 = -1 x + 4 = ± √ ―― -1 x = -4 ± iThere are two non-real solutions: -4 + i and -4 - i.

x 2 + 10x = 7 x 2 + 10x + 25 = 7 + 25 (x + 5) 2 = 32 x + 5 = ± √

_ 32

x = -5 ± 4 √_ 2

There are two non-real solutions: -5 + 4 √

_ 2 and -5 - 4 √

_ 2 .




A2_MNLESE385894_U2M03L3 142 6/8/15 1:22 PM

DIFFERENTIATE INSTRUCTION

Cognitive StrategiesSome students have trouble completing the square because there are so many steps. Show them how to break the process into three parts: (1) Get the equation into the form needed for completing the square. (2) Complete the square.(3) Finish the solution by taking square roots of both sides and simplifying the results.

When students make errors, analyze their work carefully to see what part of the process is giving them trouble, and give them extra practice on that part of the process.

EXPLAIN 2 Identifying Whether Solutions are Real or Non-real

QUESTIONING STRATEGIESDoes the discriminant give the solution of a quadratic equation? Explain. No, it gives the

number of solutions and type of solution, but it does not give the actual solution.

AVOID COMMON ERRORSRemind students that they must write the quadratic equation in standard form before applying the quadratic formula.

CONNECT VOCABULARY Review vocabulary related to quadratic functions, such as discriminant and real numbers, by having students label the parts of a quadratic function written in various forms.



© H

oug

hton

Mif

flin

Har

cour

t Pub

lishi

ng

Com

pan

y • I

mag

e C

red

its:

©D

avid

Bu

rton

/Ala

my

A person wants to create a vegetable garden and keep the rabbits out by enclosing it with 100 feet of fencing. The area of the garden is given by the function A (w) = w (50 - w) where w is the width (in feet) of the garden. Can the garden have an area of 700 ft 2 ?

Set A (w) equal to 700. w (50 - w) =

Multiply on the left side. 50w - w 2 =

Subtract 700 from both sides. - w 2 + 50w - = 0

Find the value of the discriminant.

Because the discriminant is [positive/zero/negative], the equation has [two real/one real/two non-real] solutions, so the garden [can/cannot] have an area of 700 ft 2 .

Your Turn

Answer the question by writing an equation and determining if the solutions are real or non-real.

7. A hobbyist is making a toy sailboat. For the triangular sail, she wants the height h (in inches) to be twice the length of the base b (in inches). Can the area of the sail be 10 i n 2 ?

Explain 3 Finding Complex Solutions Using the Quadratic Formula

When using the quadratic formula to solve a quadratic equation, be sure the equation is in the form a x 2 + bx + c = 0.

Example 3 Solve the equation using the quadratic formula. Check a solution by substitution.

-5 x 2 - 2x - 8 = 0

Write the quadratic formula. x = -b ± ――― b 2 - 4ac __ 2a

Substitute values. = - (-2) ±

――――――― (-2) 2 - 4 (-5) (-8) ___

2 (-5)

Simplify. = 2 ± ―― -156 __ -10 = 1 ± i ― 39 _ -5

700

700

700

50 2 - 4 (-1) (-700) = 2500 -2800 = -300

Write the area A of the sail as a function of b. A = 1 _ 2

b (2b) = b 2

Substitute 10 for A. 10 = b 2

Subtract 10 from both sides. 0 = b 2 - 10

Find the discriminant. 0 2 - 4 (1) (-10) = 0 + 40 = 40

Because the discriminant is positive, the equation has two real solutions, so the area of the sail can be 10 i n 2 .


DO NOT EDIT--Changes must be made through “File info”CorrectionKey=NL-B;CA-B

A2_MNLESE385894_U2M03L3 143 5/22/14 10:47 AM

LANGUAGE SUPPORT

Communicate MathStudents play “How do you know?” Give students several cards containing quadratic equations; some have real number solutions, others nonreal or complex solutions. In small groups, students draw a card and state whether the solution is real or not real. They then answer the question “How do you know?” Players take turns and sort cards into piles according to the kind of solution. By the end of the game, all players in a group must agree on card placement.

INTEGRATE MATHEMATICAL PRACTICESFocus on ReasoningMP.2 The discriminant can be used to distinguish between rational and irrational solutions. Give students several quadratic equations for which b 2 - 4ac is positive, some with rational solutions, and some with irrational solutions. Ask them to make a conjecture about how the value of the discriminant is related to whether the solutions are rational or irrational. Students should be able to explain why the solutions will be rational when the value of the discriminant is a perfect square.

EXPLAIN 3 Finding Complex Solutions Using the Quadratic Formula

QUESTIONING STRATEGIESWhy are there always two solutions to a quadratic equation that has nonreal solutions?

How are they related? Since √ ――― b 2 - 4ac is not zero, its value will be both added to and subtracted from -b in the numerator, resulting in two solutions; they are complex conjugates.

What is the general solution of a quadratic equation with only one solution? x = - b ___ 2a

143 Lesson 3 . 3


© H

oug

hton Mifflin H

arcourt Publishin

g Com

pany

So, the two solutions are - 1 _ 5 - i √_

39 _ 5 and - 1 _ 5 + i √_

39 _ 5 .

Check by substituting one of the values.

Substitute. -5 (- 1 _ 5 - i √ ― 39 _ 5 ) 2 - 2 (- 1 _ 5 - i

√ ― 39 _ 5 ) - 8 ≟ 0

Square. -5 ( 1 _ 25 + 2i √ ― 39 _ 25 - 39 _ 25 ) - 2 (- 1 _ 5 - i √ ― 39 _ 5 ) - 8 ≟ 0

Distribute. - 1 _ 5 - 2i √ ― 39 _ 5 + 39 _ 5 + 2 _ 5 + 2i √ ― 39 _ 5 - 8 ≟ 0

Simplify. 40 _ 5 - 8 ≟ 0 0 = 0

B 7 x 2 + 2x + 3 = -1

Write the equation with 0 on one side. 7 x 2 + 2x + = 0

Write the quadratic formula. x = -b ± √ ――― b 2 - 4ac __ 2a

Substitute values. = - ± √

――――――――― ( )

2

- 4 ( ) ( ) ____

2 ( )

Simplify. = - ±

――― - __ 14

= - ± i √

―― __ 14 =

- ± i √ ―― __ 7

So, the two solutions are and .

Check by substituting one of the values.

Substitute.

Square.

Distribute.

Simplify.

- 1 _ 7

+ 3i ― 3 _

7 - 1 _

7 - 3i

― 3 _ 7

4

42

2

2 16 3 3 3

108

2 7

7

7 (- 1 _ 7

+ 3i √ ― 3

_ 7

) 2

+ 2 (- 1 _ 7

+ 3i √ ― 3

_ 7

) 2

+ 4 ≟ 0

7 ( 1 _ 49

- 6i √ ― 3

_ 49

- 27 _ 49

) 2

+ 2 (- 1 _ 7

+ 3i √ ― 3

_ 7

) 2

+ 4 ≟ 0

1 _ 7

- 6i √ ― 3

_ 7

- 27 _ 7

- 2 _ 7

+ 6i √ ― 3

_ 7

+ 4 ≟ 0

- 28 _ 7

+ 4 ≟ 0

0 = 0




A2_MNLESE385894_U2M03L3.indd 144 19/03/14 11:59 AM

AVOID COMMON ERRORSStudents may have difficulty remembering the quadratic formula. Encourage students to copy the formula and have it on hand when they are working. Caution them to write the equation in standard form before identifying the values of a, b, and c to be used in the formula.

INTEGRATE MATHEMATICAL PRACTICESFocus on CommunicationMP.3 You may wish to point out that quadratic equations always have two roots. However, when the value of the discriminant is 0, the two roots happen to be the same. In this case, the quadratic is said to have a double root.



© H

oug

hton

Mif

flin

Har

cour

t Pub

lishi

ng

Com

pan

y

Your Turn

Solve the equation using the quadratic formula. Check a solution by substitution.8. 6 x 2 - 5x - 4 = 0 9. x 2 + 8x + 12 = 2x

Elaborate

10. Discussion Suppose that the quadratic equation a x 2 + bx + c = 0 has p + qi where q ≠ 0 as one of its solutions. What must the other solution be? How do you know?

11. Discussion You know that the graph of the quadratic function ƒ (x) = a x 2 + bx + c has the vertical line x = - b __ 2a as its axis of symmetry. If the graph of ƒ (x) crosses the x-axis, where do the x-intercepts occur relative to the axis of symmetry? Explain.

12. Essential Question Check-In Why is using the quadratic formula to solve a quadratic equation easier than completing the square?

x = -b ± √ ――― b 2 - 4ac

__ 2a

= - (-5) ± √ ―――――― (-5) 2 - 4 (6) (-4)

___ 2 (6)

= 5 ± √ ―― 121

_ 12

= 5 ± 11 _ 12

So, the solutions are 5 + 11

_ 12

= 4 _ 3

and 5 - 11 _ 12

= - 1 _ 2

.

Check

6 ( 4 _ 3

) 2

- 5 ( 4 _ 3

) - 4 ≟ 0

32 _ 3

- 20 _ 3

- 4 ≟ 0

0 = 0

x 2 + 6x + 12 = 0

x = -b ± √ ――― b 2 - 4ac

__ 2a

= - (-6) ± √ ―――――― (6) 2 - 4 (1) (12)

___ 2 (1)

= -6 ± √ ―― -12

__ 2

= -6 ± 2i √ ― 3

__ 2

= -3 ± i √ ― 3 So, the solutions are = -3 + i √ ― 3 and = -3 - i √ ― 3 .Check (-3 + i √ ― 3 )

2 + 8 (-3 + i √ ― 3 ) + 12 ≟ 2 (-3 + i √ ― 3 )

6 - 6i √ ― 3 - 24 + 8i √ ― 3 + 12 ≟ -6 + 2i √ ― 3 -6 + 2i √ ― 3 = -6 + 2i √ ― 3

The other solution must be p − qi. The radical √ ――― b 2 - 4ac in the quadratic formula

produces imaginary numbers when b 2 - 4ac < 0. Since √ ――― b 2 - 4ac is both added to and

subtracted from –b in the numerator of the quadratic formula, one solution will have the

form p + qi, and the other will have the form p − qi.

The x-intercepts are the solutions of f (x) = 0, which are x = -b ± √ ――― b 2 - 4ac ___________ 2a by the

quadratic formula. Writing the x-intercepts as x = - b __ 2a ± √ ――― b 2 - 4ac ________ 2a shows that

the x-intercepts are the same distance, √ ――― b 2 - 4ac

________ 2a , away from the axis of symmetry,

with one x-intercept on each side of the line: x = - b __ 2a - √ ――― b 2 - 4ac ________ 2a on one side and

x = - b __ 2a + √ ――― b 2 - 4ac ________ 2a on the other side.

The quadratic formula is the result of completing the square on the general

quadratic equation a x 2 + bx + c = 0. As long as any particular equation is in the form

a x 2 + bx + c = 0, you can simply substitute the values of a, b, and c into the quadratic

formula and obtain the solutions of the equation.



A2_MNLESE385894_U2M03L3 145 10/17/14 4:53 PM

ELABORATE INTEGRATE MATHEMATICAL PRACTICESFocus on Critical ThinkingMP.3 Emphasize that choosing which method to use to solve a quadratic equation is as important as being able to use each method. Have students discuss when each method might be preferred.

AVOID COMMON ERRORSStudents may sometimes make a mistake in sign when calculating the discriminant, particularly when the quantity 4ac is less than 0. Remind them that subtracting a negative number is the same as adding the opposite, or positive, number. If a and c are opposite signs, the discriminant will always be positive.

SUMMARIZE THE LESSONWhen does a quadratic equation have nonreal solutions, and how do you find them? When

the value of the discriminant is negative, the quadratic equation will have two nonreal solutions. You find the solutions by using the quadratic formula to solve the equation, and then writing the solutions as a pair of complex conjugates of the form a±bi.

145 Lesson 3 . 3


© H

oug

hton Mifflin H

arcourt Publishin

g Com

pany

• Online Homework• Hints and Help• Extra Practice

Evaluate: Homework and Practice

1. The graph of ƒ (x) = x 2 + 6x is shown. Use the graph to determine how many real solutions the following equations have: x 2 + 6x + 6 = 0, x 2 + 6x + 9 = 0, and x 2 + 6x + 12 = 0. Explain.

2. The graph of ƒ (x) = - 1 _ 2 x 2 + 3x is shown. Use the graph to determine how many real solutions the following equations have: - 1 _ 2 x 2 + 3x - 3 = 0, - 1 _ 2 x 2 + 3x - 9 _ 2 = 0, and - 1 _ 2 x 2 + 3x - 6 = 0. Explain.

Solve the equation by completing the square. State whether the solutions are real or non-real.

3. x 2 + 4x + 1 = 0 4. x 2 + 2x + 8 = 0

0-4-8-12

y4

-4

-12

4

x

0 2 31

y6

23

1

45

4 5 6 7

x

For each equation, subtract the constant from both sides to obtain these equations: x 2 + 6x = -6, x 2 + 6x = -9, and x 2 + 6x = -12.

The graph of g (x) = -6 intersects the graph of f (x) twice, so the equation x 2 + 6x + 6 = 0 has two real solutions. The graph of g (x) = -9 intersects the graph of f (x) once, so the equation x 2 + 6x + 9 = 0 has one real solution. The graph of g (x) = -12 doesn’t intersect the graph of f (x) , so the equation x 2 + 6x + 12 = 0 has no real solutions.

For each equation, subtract the constant from both sides to

obtain these equations: - 1 _ 2 x 2 + 3x = 3, - 1 _ 2 x 2 + 3x = 9 _ 2 , and

- 1 _ 2 x 2 + 3x = 6. The graph of g (x) = 3 intersects the graph of

f (x) twice, so the equation - 1 _ 2 x 2 + 3x - 3 = 0 has two real

solutions. The graph of g (x) = 9 _ 2 intersects the graph of f (x)

once, so the equation - 1 _ 2 x 2 + 3x - 9 _ 2 = 0 has one real solution.

The graph of g (x) = 6 doesn’t intersect the graph of f (x) , so the

equation - 1 _ 2 x 2 + 3x - 6 = 0 has no real solutions.

x 2 + 4x = -1

x 2 + 4x + 4 = -1 + 4

(x + 2) 2 = 3

x + 2 = ± √ ― 3

x = -2 ± √ ― 3

two real solutions: -2 + √ ― 3 and -2 - √ ― 3 .

x 2 + 2x = -8

x 2 + 2x + 1 = -8 + 1

(x + 1) 2 = -7

x + 1 = ± √ ―― -7

x + 1 = ± i √ ― 7

x = -1 ± i √ ― 7

two non-real solutions: -1 + i √ ― 7 and -1 - i √ ― 7 .




A2_MNLESE385894_U2M03L3 146 5/22/14 10:47 AMExercise Depth of Knowledge (D.O.K.)COMMONCORE Mathematical Practices

1–8 1 Recall of Information MP.2 Reasoning

9–16 1 Recall of Information MP.3 Logic

17–20 2 Skills/Concepts MP.2 Reasoning

21 1 Recall of Information MP.3 Logic

22 3 Strategic Thinking MP.3 Logic

23–24 3 Strategic Thinking MP.4 Modeling

EVALUATE

ASSIGNMENT GUIDE

Concepts and Skills Practice

ExploreInvestigating Real Solutions of Quadratic Equations

Exercises 1–2

Example 1Finding Complex Solutions by Completing the Square

Exercises 3–8

Example 2Identifying Whether Solutions are Real or Non-real

Exercises 9–16

Example 3Finding Complex Solutions Using the Quadratic Formula

Exercises 17–20

CONNECT VOCABULARY What information does the value of the discriminant give about a quadratic equation? The value of the discriminant indicates the number and types of roots.



© H

oug

hton

Mif

flin

Har

cour

t Pub

lishi

ng

Com

pan

y

5. x 2 - 5x = -20 6. 5 x 2 - 6x = 8

7. 7 x 2 + 13x = 5 8. - x 2 - 6x - 11 = 0

Without solving the equation, state the number of solutions and whether they are real or non-real.9. -16 x 2 + 4x + 13 = 0 10. 7 x 2 - 11x + 10 = 0

11. - x 2 - 2 _ 5 x = 1 12. 4 x 2 + 9 = 12x

x 2 - 5x + 25 __ 4 = -20 + 25 __ 4

(x - 5 _ 2 ) 2

= - 55 __ 4

x - 5 _ 2 = ± √ ―― - 55 __ 4

x - 5 _ 2 = ± i √ ― 55

____ 2

x = 5 _ 2 ± i √ ― 55

____ 2

two non-real solutions:

5 _ 2 + i √ ― 55

____ 2 and 5 _ 2 - i √ ― 55

____ 2 .

x 2 - 1.2x = 1.6

x 2 - 1.2x + 0.36 = 1.6 + 0.36

(x - 0.6) 2 = 1.96

x - 0.6 = ± √ ―― 1.96

x - 0.6 = ±1.4

x = 0.6 ± 1.4

two real solutions: 2 and -0.8.

x 2 + 13 __ 7 x = 5 _ 7

x 2 + 13 __ 7 x + 169 ___ 196 = 5 _ 7 + 169 ___ 196

(x + 13 __ 14 ) 2

= 309 ___ 196

x + 13 __ 14 = ± √ ―― 309 ___ 196

x + 13 __ 14 = ± √ ―― 309

_____ 14

x = - 13 __ 14 ± √ ―― 309

_____ 14

two real solutions:

-13 + √ ―― 309 _________ 14 and -13 - √ ―― 309

_________ 14 .

x 2 + 6x + 11 = 0

x 2 + 6x + 9 = -11 + 9

(x + 3) 2 = -2

x + 3 = ± √ ―― -2

x + 3 = ± i √ ― 2

x = -3 ± i √ ― 2

two non-real solutions: -3 + i √ ― 2 and -3 - i √ ― 2 .

Find the discriminant.

4 2 - 4 (-16) (13) = 16 + 832 = 848

Because the discriminant is positive, the equation has two real solutions.


(-11) 2 - 4 (7) (10) = 121 - 280 = -159

Because the discriminant is negative, the equation has two non-real solutions.

- x 2 - 2 _ 5 x - 1 = 0


(- 2 _ 5 ) 2

- 4 (-1) (-1) = 4 __ 25 - 4 = - 96 __ 25

Because the discriminant is negative, the equation has two non-real solutions.

4 x 2 - 12x + 9 = 0


(-12) 2 - 4 (4) (9) = 144 - 144 = 0

Because the discriminant is zero, the equation has one real solution.




VISUAL CUESIf students have difficulty evaluating the discriminant, have them organize the variables in a table. Ask students to create a table for each of the variables (a, b, c, b 2 , 4ac, b 2 - 4ac) and have them predict the number and type of solutions based on the variables they list in the table.

147 Lesson 3 . 3


© H

oug

hton Mifflin H

arcourt Publishin

g Com

pany • Im

age C

redits: ©

Aflo Foto

Ag

ency/A

lamy

Answer the question by writing an equation and determining whether the solutions of the equation are real or non-real.

13. A gardener has 140 feet of fencing to put around a rectangular vegetable garden. The function A (w) = 70w - w 2 gives the garden’s area A (in square feet) for any width w (in feet). Does the gardener have enough fencing for the area of the garden to be 1300 ft 2 ?

14. A golf ball is hit with an initial vertical velocity of 64 ft/s. The function h (t) = -16t 2 + 64t models the height h (in feet) of the golf ball at time t (in seconds). Does the golf ball reach a height of 60 ft?

15. As a decoration for a school dance, the student council creates a parabolic arch with balloons attached to it for students to walk through as they enter the dance. The shape of the arch is modeled by the equation y = x (5 - x) , where x and y are measured in feet and where the origin is at one end of the arch. Can a student who is 6 feet 6 inches tall walk through the arch without ducking?

Write an equation by setting A (w) equal to 1300. Then rewrite the equation

with 0 on one side. 70w - w 2 = 1300

-w 2 + 70w − 1300 = 0

Find the discriminant. 70 2 − 4 (−1) (−1300) = 4900 − 5200 = − 300

Because the discriminant is negative, the equation has two non-real

solutions, so the gardener does not have enough fencing.

Write an equation by setting h (t) equal to 60. Then rewrite the

equation with 0 on one side. -16t 2 + 64t = 60

-16t 2 + 64t - 60 = 0

4t 2 - 16t + 15 = 0

Find the discriminant. (-16) 2 − 4 (4) (15) = 256 − 240 = 16

Because the discriminant is positive, the equation has two real

solutions, so the golf ball does reach a height of 60 ft.

Write an equation by setting y equal to 6.5. Then rewrite the equation with 0 on

one side. x (5 - x) = 6.5

5x - x 2 =6.5

-x 2 + 5x - 6.5 =0

Find the discriminant. 5 2 -4 (-1) (-6.5) = 25 - 26 = -1

Because the discriminant is negative, the equation has two non-real solutions, so a

student who is 6 feet 6 inches tall cannot walk through the arch without ducking.




A2_MNLESE385894_U2M03L3 148 10/17/14 4:56 PM



© H

oug

hton

Mif

flin

Har

cour

t Pub

lishi

ng

Com

pan

y

16. A small theater company currently has 200 subscribers who each pay $120 for a season ticket. The revenue from season-ticket subscriptions is $24,000. Market research indicates that for each $10 increase in the cost of a season ticket, the theater company will lose 10 subscribers. A model for the projected revenue R (in dollars) from season-ticket subscriptions is R (p) = (120 + 10p) (200 - 10p) , where p is the number of $10 price increases. According to this model, is it possible for the theater company to generate $25,600 in revenue by increasing the price of a season ticket?

Solve the equation using the quadratic formula. Check a solution by substitution.

17. x 2 - 8x + 27 = 0 18. x 2 - 30x+ 50 = 0

Write an equation by setting R (p) equal to 25,600. Then rewrite the equation with 0 on

one side. (120 + 10p) (200 - 10p) = 25,600

-100p 2 + 800p + 24, 000 = 25,600

-100p 2 + 800p - 1600 = 0

p 2 - 8p + 16 = 0

Find the discriminant. (-8) 2 - 4 (1) (16) = 64 - 64 = 0

Because the discriminant is zero, the equation has one real solution, so it is possible to

generate $25,600 in revenue by increasing the price of a season ticket.

x = -b ± √ ――― b 2 - 4ac __

2a

= - (-8) ± √ ―――――― (-8) 2 - 4 (1) (27)

___ 2 (1)

= 8 ± √ ―― -44

_ 2

= 8 ± 2i √ ― 11

_ 2

= 4 ± i √ ― 11

So, the solutions are

4 + i √ ― 11 and 4 - i √ ― 11 .

Check

(4 + i √ ― 11 ) 2 - 8 (4 + i √ ― 11 ) + 27 ≟ 0

5 + 8i √ ― 11 - 8 (4 + i √ ― 11 ) + 27 ≟ 0

5 + 8i √ ― 11 - 32 - 8i √ ― 11 + 27 ≟ 0

5 - 32 + 27 ≟ 0

0 = 0

x = -b ± √ ――― b 2 - 4ac __

2a

= - (-30) ± √ ――――――― (-30) 2 - 4 (1) (50)

___ 2 (1)

= 30 ± √ ―― 700

__ 2

= 30 ± 10 √ ― 7

__ 2

= 15 ± 5 √ ― 7

So, the solutions are

15 + 5 √ ― 7 and 15 - 5 √ ― 7 .

Check

(15 + 5 √ ― 7 ) 2 - 30 (15 + 5 √ ― 7 ) + 50 ≟ 0

400 + 150 √ ― 7 - 30 (15 + 5 √ ― 7 ) + 50 ≟ 0

400 + 150 √ ― 7 - 450 - 150 √ ― 7 + 50 ≟ 0

400 - 150 + 50 ≟ 0

0 = 0



A2_MNLESE385894_U2M03L3 149 5/22/14 10:47 AM

PEER-TO-PEER DISCUSSIONAsk students to discuss with a partner how the graphs of the following three parabolas would look: a parabola with two real solutions, a parabola with one real solution, and a parabola with two nonreal solutions. Students should say that a parabola with two solutions will have two x-intercepts, and the parabola will open from the vertex toward the x-axis; that a parabola with one solution will have one x-intercept with the vertex on the x-axis; and that a parabola with two nonreal solutions will open from the vertex away from the axis and have no x-intercept.

149 Lesson 3 . 3


© H

oug

hton Mifflin H

arcourt Publishin

g Com

pany

19. x + 3 = x 2 20. 2 x 2 + 7 = 4x

21. Place an X in the appropriate column of the table to classify each equation by the number and type of its solutions.

Equation Two Real Solutions

One Real Solution

Two Non-Real Solutions

x 2 - 3x + 1 = 0

x 2 - 2x + 1 = 0

x 2 - x +1 = 0

x 2 + 1 = 0

x 2 + x + 1 = 0

x 2 + 2x + 1 = 0

x 2 + 3x + 1 = 0

Rewrite the equation with 0 on one side.

x 2 - x -3 = 0 Use the quadratic formula.

x = -b ± √ ――― b 2 - 4ac

__ 2a

= - (-1) ± √ ―――――― (-1) 2 - 4 (1) (-3)

___ 2 (1)

= 1 ± √ ― 13

_ 2

So, the two solutions are

1 + √ ― 13 ______ 2 and 1 - √ ― 13

______ 2 .

Check

( 1 + √ ― 13

_ 2

) 2

- ( 1 + √ ― 13

_ 2

) - 3 ≟ 0

14 + 2 √ ― 13

__ 4

- ( 1 + √ ― 13

_ 2

) - 3 ≟ 0

14 + 2 √ ― 13

__ 4

- ( 2 + 2 √ ― 13

_ 4

) - 3 ≟ 0

12 _ 4

- 3 ≟ 0

0 = 0

2x 2 - 4x + 7 = 0

x = -b ± √ ――― b 2 - 4ac __

2a

= - (-4) ± √ ―――――― (-4) 2 - 4 (2) (7)

___ 2 (2)

= 4 ± √ ―― -40

__ 4

= 4 ± 2i √ ― 10

__ 4

= 2 ± i √ ― 10

_ 2

So, the two solutions are

1 + i √ ― 10

____ 2 and 1 - i √ ― 10

____ 2 .

Check.

2 (1+ i √ ― 10

_ 2

) 2

- 4 (1 + i √ ― 10

_ 2

) + 7 ≟ 0

2 (- 3 _ 2

+ i √ ― 10 ) - 4 (1 + i √ ― 10

_ 2

) + 7 ≟ 0

-3 + 2i √ ― 10 - 4 - 2i √ ― 10 + 7 ≟ 0

-3 - 4 + 7 ≟ 0

0 = 0

X

X

X

X

X

X

X





AVOID COMMON ERRORSStudents need to be careful to avoid making sign errors when completing the square. Point out that when the rule representing vertex form is simplified, the result should be the original rule written in standard form. Students can use this fact to perform a quick check of the reasonableness of their results, and in order to catch any sign errors they may have made.



© H

oug

hton

Mif

flin

Har

cour

t Pub

lishi

ng

Com

pan

y

H.O.T. Focus on Higher Order Thinking

22. Explain the Error A student used the method of completing the square to solve the equation -x 2 + 2x - 3 = 0. Describe and correct the error.

-x 2 + 2x - 3 = 0

-x 2 + 2x = 3

-x 2 + 2x + 1 = 3 + 1

(x + 1) 2 = 4

x + 1 = ± √_

4

x + 1 = ±2

x = -1 ± 2

So, the two solutions are -1 + 2 = 1 and -1 - 2 = -3.

23. Make a Conjecture Describe the values of c for which the equation x 2 + 8x + c = 0 has two real solutions, one real solution, and two non-real solutions.

24. Analyze Relationships When you rewrite y = ax 2 + bx + c in vertex form by completing the square, you obtain these coordinates for the vertex: (- b __ 2a , c - b 2 __ 4a ) . Suppose the vertex of the graph of y = ax 2 + bx + c is located on the x-axis. Explain how the coordinates of the vertex and the quadratic formula are in agreement in this situation.

The student did not divide both sides by –1 first to make the coefficient of the x 2 -term be 1. The correct solution is as follows.

x 2 - 2x + 3 = 0 x 2 - 2x = -3

x 2 - 2x + 1 = -3 + 1

(x -1) 2 = -2

x - 1 = ± √ ―― -2

x - 1 = ± i √ ― 2

x = 1 ± i √ ― 2

So, the two solutions are 1 + i √ ― 2 and 1 - i √ ― 2 .

Find the value of the discriminant.

b 2 - 4ac = 8 2 - 4 (1) c = 64 - 4c

The equation has two real solutions when the discriminant is positive, so solving 64 - 4c > 0 for c gives c < 16. The equation has one real solution when the discriminant is zero, so solving 64 - 4c = 0 for c gives c = 16. The equation has two non-real solutions when the discriminant is negative, so solving 64 - 4c < 0 for c gives c > 16.

When the vertex is on the x-axis, the y-coordinate of the vertex must be 0,

so c - b 2 __ 4a = 0, which can be rewritten as b 2 - 4ac = 0. When you set y equal

to 0 in y = ax 2 + bx + c and solve for x, you get one real solution, namely,

x = - b __ 2a , which is the x-coordinate of the vertex.




JOURNALHave students summarize how to use the discriminant to help solve any quadratic equation. Have them include examples of quadratic equations with one or two real solutions and with two nonreal solutions.

151 Lesson 3 . 3


© H

oug

hton Mifflin H

arcourt Publishin

g Com

pany

Lesson Performance Task

Matt and his friends are enjoying an afternoon at a baseball game. A batter hits a towering home run, and Matt shouts, “Wow, that must have been 110 feet high!” The ball was 4 feet off the ground when the batter hit it, and the ball came off the bat traveling vertically at 80 feet per second.

a. Model the ball’s height h (in feet) at time t (in seconds) using the projectile motion model h (t) = -16 t 2 + v 0 t + h 0 where v 0 is the projectile’s initial vertical velocity (in feet per second) and h 0 is the projectile’s initial height (in feet). Use the model to write an equation based on Matt’s claim, and then determine whether Matt’s claim is correct.

b. Did the ball reach a height of 100 feet? Explain.

c. Let h max be the ball’s maximum height. By setting the projectile motion model equal to h max , show how you can find h max using the discriminant of the quadratic formula.

d. Find the time at which the ball reached its maximum height.

a. The ball’s height h at time t is given by h (t) = -16t 2 + 80t + 4. Matt’s claim is that h (t) = 110 at some time t. Applying the discriminant of the quadratic formula to the equation -16t 2 + 80t + 4 = 110, or -16t 2 + 80t - 106 = 0, gives b 2 - 4ac = 80 2 - 4 (-16) (-106) = 6400 - 6784 = -384. Since the discriminant is negative, there are no real values of t that solve the equation, so Matt’s claim is incorrect.

b. For the ball to reach of height of 100 feet, h (t) must equal 100. Applying the discriminant of the quadratic formula to the equation -16t 2 + 80t + 4 = 100, or -16t 2 + 80t - 96 = 0, gives b 2 - 4ac = 80 2 - 4 (-16) (-96) = 6400 - 6144 = 256. Since the discriminant is positive, there are two real values of t that solve the equation, so the ball did reach a height of 100 feet at two different times (once before reaching its maximum height and once after).

c. Setting h (t) equal to h max gives -16 t 2 + 80t + 4 = h max = 0, or -16 t 2 + 80t + 4 - h max = 0. Since the maximum height occurs for a single real value of t, the discriminant of the quadratic equation must equal 0.

b 2 - 4ac = 0 80 2 - 4 (-16) (4 - h max ) = 0 6400 + 64 (4 - h max ) = 0 64 (4 - h max ) = -6400 4 - h max = -100 - h max = -104 h max = 104 So, the ball reached a maximum height of 104 feet. d. Solve the equation -16t 2 + 80t + 4 = 104, or -16t 2 + 80t - 100 = 0, using the quadratic

formula. You already know that the discriminant is 0 when the ball reached its maximum

height, so t = -80 ± √ ― 0 ________

2 (-16) = -80 ___ -32 = 2.5. So, the ball reached its maximum height 2.5 seconds

after it was hit.


DO NOT EDIT--Changes must be made through “File info” CorrectionKey=NL-C;CA-C

A2_MNLESE385894_U2M03L3 152 6/8/15 1:23 PM

EXTENSION ACTIVITY

On the moon, the force of gravity is 1 _ 6 Earth’s gravity, so the equation for simple projectile motion for a ball hit at a height of 4 feet above the ground is y = - 16 _ 6 t 2 + vt + 4. Have students determine if a baseball hit upward traveling at an initial vertical velocity of v = 80 ft/s on the moon reaches a height of 200 feet. They should find the real solutions. t = 2.7 and 27.3 s, so the ball does reach a height of 200 feet.

AVOID COMMON ERRORSStudents may sometimes make a mistake in sign when calculating the discriminant, particularly when the quantity 4ac is less than 0. Remind them that subtracting a negative number is the same as adding the opposite, or positive, number. If a and c have opposite signs, the discriminant will always be positive.

INTEGRATE TECHNOLOGYStudents can use a graphing utility to graph a parabola and find the maximum value.

QUESTIONING STRATEGIESHow can the symmetry of a parabola help you to find the maximum or minimum if you

know two different points on the graph with the same y-coordinate? The x-coordinate of the maximum or minimum will be halfway between the x-coordinates of the two points on the graph.

Scoring Rubric2 points: Student correctly solves the problem and explains his/her reasoning.1 point: Student shows good understanding of the problem but does not fully solve or explain his/her reasoning.0 points: Student does not demonstrate understanding of the problem.



Documents

CorrectionKey=NL-A;CA-A CorrectionKey=NL-C;CA-C 3 . 3 DO